Answer:
It should take around 15 minutes
Step-by-step explanation:
The equation gives the height of the ball. That is, h is the height of the ball. t is the time. Since we are looking for the time at which the height is 8 (h=8), we need to set the equation equal to 8 and solve for t. We do this as follows:




This is a quadratic equation and as it is set equal to 0 we can solve it using the quadratic formula. That formula is:

You might recall seeing this as "x=..." but since our equation is in terms of t we use "t-=..."
In order to use the formula we need to identify a, b and c.
a = the coefficient (number in front of)

= 16.
b = the coefficient of t = -60
c = the constant (the number that is by itself) = 7
Substituting these into the quadratic formula gives us:



As we have "plus minus" (this is usually written in symbols with a plus sign over a minus sign) we split the equation in two and obtain:

and

So the height is 8 feet at t = 3.63 and t=.12
It should make sense that there are two times. The ball goes up, reaches it's highest height and then comes back down. As such the height will be 8 at some point on the way up and also at some point on the way down.
The Z-score (z) is calculated according to the formula: z = (x - μ) / σ
Where: x is the raw score value, μ is the mean of the population, σ is the standard deviation of the population.
x = 80, μ = 5, σ = 4
z = (80 - 5) / 4
z = 75 / 4 = 18.7500
Z-score is 18.7500
6 goes in to 906 151 times